Questions about homotopy lifting theorem. I’m reading the following material and would like to ask some clarification.

Theorem B8.5 (Homotopy lifting). Let $p: \tilde{X} \rightarrow X$ be a covering, and $H: Y \times I \rightarrow X$ be a homotopy between maps $f_{0}, f_{1}: Y \rightarrow X .$ Given any lift $\tilde{f}_{0}: Y \rightarrow \tilde{X}$ of $f_{0},$ there is $a$
unique lift $\tilde{H}: Y \times X \rightarrow \tilde{X}$ of $H$ agreeing with $\tilde{f}_{0}$ on $Y \times\{0\}$


Proof. For each $y \in Y,$ there is a unique lift of the path $H(y, \cdot): I \rightarrow X$ starting at $\tilde{f}_{0}(y) .$ Collecting these lifts, we get a possibly discontinuous function $\tilde{H}: Y \times I \rightarrow \tilde{X} ;$ we just need to show that this $\tilde{H}$ is continuous. Fixing any $y \in Y$ it suffices to find a neighborhood $N \ni y$ such that $\tilde{H}$ is continuous on $N \times I$.
Consider a cover $\left\{U_{\alpha}\right\}$ of $X$ by trivializing neighborhoods for the covering $p: \tilde{X} \rightarrow X .$ The pre-images $H^{-1} U_{\alpha}$ cover $Y \times I .$ By the corollary to the Lebesgue lemma, we can find a neighborhood $N$ of $y$ and a partition $\left\{t_{i}\right\}$ of $I$ such that each $H\left(N \times\left[t_{i-1}, t_{i}\right]\right)$ is contained in a trivializing neighborhood that we call $U_{i}$.
Start with $\tilde{H}=\tilde{f}_{0}$ on $N \times\left\{t_{0}\right\}$ and proceed by induction on $i=1, \ldots, n .$ If we know $\tilde{H}$ on $N \times\left\{t_{i-1}\right\},$ then we look at the trivialization of the covering over $U_{i} .$ On $N \times\left\{t_{i-1}\right\},$ the lift $\tilde{H}$ is given by a continuous map to the fiber $F ;$ since the interval $\left[t_{i-1}, t_{i}\right]$ is connected, the only way to extend the lift to $N \times\left[t_{i-1}, t_{i}\right]$ is to make it constant in $t .$ Thus $\tilde{H}$ is continuous on $N \times\left[t_{i-1}, t_{i}\right],$ and in particular on $N \times\left\{t_{i}\right\}$, which is what we need to continue the induction. Once we reach $t_{n}=1,$ we see that $\tilde{H}$ is continuous on all of $N \times I$ as claimed.

Here’s the statement of the corollary to Lebesgue’s lemma:

Corollary B7.3. Suppose $Y$ is a topological space, $\left\{U_{\alpha}\right\}$ is an open cover of $Y \times I$, and $y \in Y$ is any point. Then there exists a neighborhood $N$ с $Y$ of $y$ and a finite partition $0=t_{0}<t_{1}<\cdots<t_{n}=1$ of I such that each $N \times\left[t_{i-1}, t_{i}\right]$
is contained in some $U_{\alpha}$.

My questions are as follows:

*

*Is it absolutely necessary that the interval $I$ in $N \times I$ is partitioned into several intervals $[t_{i-1},t_i]$? Shouldn’t $H(N \times I)$ already fit into some $U_\alpha$? The reason for this question is because I’m picturing the pre-images $H^{-1}U_\alpha$ as $U_\alpha \times I$ (so they are like “cylinders”, and $N \times I$ would just fit into the intersection of some of those).

*I’m not clear about the last part of the argument, in particular this statement:


On $N \times\left\{t_{i-1}\right\},$ the lift $\tilde{H}$ is given by a continuous map to the fiber $F ;$ since the interval $\left[t_{i-1}, t_{i}\right]$ is connected, the only way to extend the lift to $N \times\left[t_{i-1}, t_{i}\right]$ is to make it constant in $t .$

It’d be great if someone could explain. (Also as a side note, another argument is that the lift is given by a locally constant function to $F$, hence the conclusion. But in an earlier part of the text it specifies that the function to $F$ is locally constant if the fiber is discrete. How does it extend to this setting, where the fiber is $I$?)
 A: Here is a simple example which exhibits that the answers to your questions in 1 are "no".
Let $Y = I$, and let $U_\alpha$, for each point $\alpha = (y,t) \in Y \times I$, be the box neighborhood
$$U_\alpha = (I \times I) \cap \bigl((y-.001,y+.001) \times (t-.001,t+.001)\bigr)
$$
You can see, for example, that its going to take a few hundred thousand separate elements of the collection $\{U_\alpha\}$ just to cover $Y \times I = I \times I$, because each $U_\alpha$ has area $\le .002 \times .002 = .000004$ and $I \times I$ has area $1$.
Now pick any $y \in Y$, and any neighborhood $N \subset Y$ of $y$. Then there does not exist any $\alpha$ such that $y \times I \subset U_\alpha$, and therefore there also does not exist any $\alpha$ such that $N \times I \subset U_\alpha$. I fact, you can see that its going to at least a five hundred separate $U_\alpha$'s in order to cover just $y \times I$, because each $U_\alpha$ has height $\le .003$ whereas $y \times I$ has height $1$. So it will also take at least five hundred separate $U_\alpha$'s to cover $N \times I$.
Perhaps what sent you astray is in thinking that $H^{-1} U_\alpha$ has the form $U_\alpha \times I \subset Y \times I$. That doesn't even make sense, $U_\alpha$ is a subset of $X$, not of $Y$. (I also worry that the clashing uses of $U_\alpha$ in Theorem B8.5 and Corollary B7.3 could be confusing).
Regarding 2, the phrase "...is to make it constant in $t$" is ambiguous: What is it? To make this clearer, you should read that phrase as "is to make the lift to the fiber constant in $t$". Keep in mind, the fiber $F$ that we are working with in this passage of the proof is the one in the formula $p^{-1}(U_\alpha) \approx U_\alpha \times F$ which comes from the definition of a "trivializing neighborhood". What we are trying to construct is the continuous function $\tilde H : N \times [t_{i-1},t_i] \to U_\alpha \times F$. Since $F$ is discrete, every continuous function to $F$ from a connected space (such as $[t_{i-1},t_i]$) is constant.
